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Câu 2 :
a) \(\frac{1}{4}\) + \(\frac{1}{3}\): (3x) = -5
⇒ \(\frac{1}{3}\): (3x) = -5 - \(\frac{1}{4}\)
⇒ \(\frac{1}{3}\): (3x) = \(\frac{-21}{4}\)
⇒ 3x = \(\frac{1}{3}\): \(\frac{-21}{4}\)
⇒ 3x = \(\frac{-4}{63}\)
⇒ x = \(\frac{-4}{63}\):3
⇒ x = \(\frac{-4}{189}\)
Vậy x = \(\frac{-4}{189}\)
b) (3x-4) . (5x+15)=0
⇒ xảy ra 2 trường hợp 3x-4=0 ; 5x+15=0
* 3x-4 =0
⇒ 3x =0+4
⇒ 3x =4
⇒ x =4:3
⇒ Vô lý không tính được bạn nhé
* 5x+15 = 0
⇒ 5x = 0-15
⇒ 5x = -15
⇒ x = -15:5
⇒ x = -3
Vậy x ∈ ∅ và x ∈ 3
c) |2x-1| = \(\frac{11}{2}\)
⇒ xảy ra 2 trường hợp 2x-1 = \(\frac{11}{2}\); 2x-1 = \(\frac{-11}{2}\)
* 2x-1=\(\frac{11}{2}\)
⇒ 2x = \(\frac{11}{2}\)+1
⇒ 2x = \(\frac{9}{2}\)
⇒ x = \(\frac{9}{2}\):2
⇒ x = \(\frac{9}{4}\)
* 2x-1 = \(\frac{-11}{2}\)
⇒ 2x = 1 + \(\frac{-11}{2}\)
⇒ 2x = \(\frac{-9}{2}\)
⇒ x = \(\frac{-9}{2}\):2
⇒ x = \(\frac{-9}{4}\)
Vậy x ∈ { \(\frac{9}{4}\); \(\frac{-9}{4}\)}
Câu 2:
a) \(\frac{1}{4}+\frac{1}{3}:\left(3x\right)=-5\)
\(\Leftrightarrow\frac{1}{3}\cdot\frac{1}{3x}=-5-\frac{1}{4}=-\frac{21}{4}\)
\(\Leftrightarrow\frac{1}{9x}=\frac{-21}{4}\)
\(\Leftrightarrow9x=\frac{4\cdot1}{-21}=-\frac{4}{21}\)
hay \(x=-\frac{4}{21}:9=-\frac{4}{189}\)
Vậy: \(x=-\frac{4}{189}\)
b) Ta có: \(\left(3x-4\right)\left(5x+15\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}3x-4=0\\5x+15=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}3x=4\\5x=-15\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=\frac{4}{3}\\x=-3\end{matrix}\right.\)
Vậy: \(x\in\left\{\frac{4}{3};-3\right\}\)
c) Ta có: \(\left|2x-1\right|=\frac{11}{2}\)
\(\Leftrightarrow\left[{}\begin{matrix}2x-1=\frac{11}{2}\\2x-1=-\frac{11}{2}\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}2x=\frac{11}{2}+1=\frac{13}{2}\\2x=-\frac{11}{2}+1=-\frac{9}{2}\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x=\frac{13}{2}:2=\frac{13}{4}\\x=-\frac{9}{2}:2=-\frac{9}{4}\end{matrix}\right.\)
Vậy: \(x\in\left\{\frac{13}{4};\frac{-9}{4}\right\}\)
Câu 3:
a) Ta có: \(1-3\cdot\left[4-30:\left(-18+3\right)\right]\)
\(=1-3\cdot\left[4-30:\left(-15\right)\right]\)
\(=1-3\cdot\left[4-\left(-2\right)\right]\)
\(=1-3\cdot6=1-18=-17\)
b) Ta có: \(\frac{5\cdot7+5\cdot\left(-4\right)}{21\cdot5}\)
\(=\frac{5\cdot\left(7-4\right)}{5\cdot21}=\frac{3}{21}=\frac{1}{7}\)
c) Ta có: \(\frac{-2}{9}+\frac{5}{4}+\left(-\frac{1}{6}\right):\frac{3}{5}+\frac{1}{18}\)
\(=-\frac{2}{9}+\frac{5}{4}-\frac{5}{18}+\frac{1}{18}\)
\(=-\frac{2}{9}+\frac{5}{4}-\frac{4}{18}\)
\(=\frac{-8}{36}+\frac{45}{36}-\frac{8}{36}=\frac{29}{36}\)
a) \(\frac{6}{7}\) và \(\frac{11}{10}\)
\(\frac{6}{7}< 1\)
\(\frac{11}{10}>1\)
\(\Rightarrow\frac{6}{7}< 1< \frac{11}{10}\Rightarrow\frac{6}{7}< \frac{11}{10}\)
b) \(\frac{-5}{17}\) và \(\frac{2}{7}\)
\(\frac{-5}{17}< 0\)
\(\frac{2}{7}>0\)
\(\Rightarrow\frac{-5}{17}< 0< \frac{2}{7}\)\(\Rightarrow\frac{-5}{17}< \frac{2}{7}\)
c) \(\frac{419}{-723}\) và \(\frac{-697}{-313}\)
\(\frac{419}{-724}< 0\)
\(\frac{-697}{-313}>0\)
\(\Rightarrow\frac{419}{-724}< 0< \frac{-697}{-313}\Rightarrow\frac{419}{-723}< \frac{-697}{-313}\)
a: \(\widehat{YOZ}=180^0-130^0=50^0\)
b: \(\widehat{MON}=\widehat{MOY}+\widehat{NOY}=\dfrac{180^0}{2}=90^0\)
Bài 1:
a) \(\dfrac{1}{n\left(n+1\right)}=\dfrac{1}{n}-\dfrac{1}{n+1}\)
Quy đồng \(VP\) ta được:
\(VP=\dfrac{1}{n}-\dfrac{1}{n+1}\)
\(\Rightarrow VP=\dfrac{n+1}{n\left(n+1\right)}-\dfrac{n}{n\left(n+1\right)}\)
\(\Rightarrow VP=\dfrac{n+1-n}{n\left(n+1\right)}=\dfrac{1}{n\left(n+1\right)}\)
\(\Rightarrow VP=VT\)
Vậy \(\forall n\in Z,n>0\Rightarrow\dfrac{1}{n\left(n+1\right)}=\dfrac{1}{n}-\dfrac{1}{n+1}\) (Đpcm)
b) \(\dfrac{1}{1.2}+\dfrac{1}{2.3}+\dfrac{1}{3.4}+...+\dfrac{1}{9.10}\)
\(=1-\dfrac{1}{2}+\dfrac{1}{2}-\dfrac{1}{3}+...+\dfrac{1}{9}-\dfrac{1}{10}\)
\(=1-\dfrac{1}{10}\)
\(=\dfrac{9}{10}\)
Bài 3:
a) \(\dfrac{1}{n}-\dfrac{1}{n+1}=\dfrac{1+1}{n\left(n+1\right)}-\dfrac{n}{n\left(n+1\right)}=\dfrac{1}{n\left(n+1\right)}\)
b) A=\(\dfrac{1}{2}.\dfrac{1}{3}+\dfrac{1}{3}.\dfrac{1}{4}+\dfrac{1}{4}.\dfrac{1}{5}+\dfrac{1}{5}.\dfrac{1}{6}+\dfrac{1}{6}.\dfrac{1}{7}+\dfrac{1}{7}.\dfrac{1}{8}+\dfrac{1}{8}.\dfrac{1}{9}\)
\(=\dfrac{1}{2}-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{4}+\dfrac{1}{4}-\dfrac{1}{5}+\dfrac{1}{5}-\dfrac{1}{6}+\dfrac{1}{6}-\dfrac{1}{7}+\dfrac{1}{7}-\dfrac{1}{8}+\dfrac{1}{8}-\dfrac{1}{9}\)
\(=\dfrac{1}{2}-\dfrac{1}{9}\)
\(=\dfrac{7}{18}\)
B=\(\dfrac{1}{30}+\dfrac{1}{42}+\dfrac{1}{56}+\dfrac{1}{72}+\dfrac{1}{90}+\dfrac{1}{110}+\dfrac{1}{132}\)
\(=\dfrac{1}{5.6}+\dfrac{1}{6.7}+\dfrac{1}{7.8}+\dfrac{1}{8.9}+\dfrac{1}{9.10}+\dfrac{1}{10.11}+\dfrac{1}{11.12}\)
\(=\dfrac{1}{5}-\dfrac{1}{12}\)
\(=\dfrac{7}{60}\)
a: \(\widehat{nOa}=180^0-60^0=120^0\)
b: \(\widehat{tOk}=\widehat{tOn}+\widehat{kOn}=\dfrac{60^0}{2}+\dfrac{120^0}{2}=90^0\)
a)3 giờ
b)15km
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